MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Use the moment of inertia of a ring about its centre, and the parallel axis theorem for the six outer rings.Explanation:The central ring has its centre on the axis.So its moment of inertia is:$$I_{\text{central}} = M R^2$$Each outer ring touches the central ring, so the distance between their centres is:$$d = R + R = 2R$$For one outer ring, its own moment of inertia about its centre is:$$I_{\text{cm}} = M R^2$$The given axis passes through the centre of the central ring, not through the outer ring's centre.Apply the parallel axis theorem:$$I_{\text{outer}} = I_{\text{cm}} + M d^2$$$$I_{\text{outer}} = M R^2 + M(2R)^2$$$$I_{\text{outer}} = M R^2 + 4 M R^2 = 5 M R^2$$There are six outer rings, so their total moment of inertia is:$$I_{\text{surrounding}} = 6 \times 5 M R^2 = 30 M R^2$$Total moment of inertia of all seven rings:$$I_{\text{total}} = I_{\text{central}} + I_{\text{surrounding}}$$$$I_{\text{total}} = M R^2 + 30 M R^2 = 31 M R^2$$Answer:$$I_{\text{total}} = 31 M R^2$$Hence, the correct option is A. $31 M R^{2}$.